help with showing completeness Let $\left\{H_n\right\}_{n=1}^\infty$ be a sequence of Hilbert spaces and let $H=\left\{\left\{x_n\right\}:x_n\in H_n, \sum ||x_n||^2<\infty \right\}$. Define the inner product as $(\left\{x_n\right\}, \left\{y_n\right\})=\sum (x_n,y_n)$ Then $H$ is complete with respect to the induced norm $\left\|x_n \right\|=(\left\{x_n\right\}, \left\{x_n\right\})^\frac{1}{2}$.
I want to consider a Cauchy sequence $\left\{ \left\{x_{i,m}  \right\}_{i=1}^\infty \right\}_{m=1}^\infty$ and use the fact that $\sum ||x_n||^2<\infty$, but here is where I run into a problem:
$\displaystyle \sum_{m=1}^\infty ( \left\{x_i  \right\}_m , \left\{x_i  \right\}_m)= \sum_{m=1}^\infty \sum_{i=1}^\infty (x_{i,m}, x_{i,m})$.
The sum above may not be necessarily finite, right? I want to show that it is finite so that this way I know that $\left\{ \left\{x_i  \right\}_m \right\}_{m=1}^\infty$ has a limit. What am I doing wrong? Thanks for your help.
 A: The proof of the completeness of $H$ is very similar to the proof of completeness of $\ell^2(\mathbb{N})$.


*

*The components in $H_n$ of the Cauchy sequence in $H$ are Cauchy sequences, and by the completeness of $H_n$ the "pointwise" limit $\xi = \{\xi_n\}_{n\in\mathbb{N}}$ exists.

*The pointwise limit $\xi$ belongs to $H$, and

*the sequence converges to $\xi$ in the norm topology on $H$.


To avoid notational confusion, let me write $x^{(m)}$ for the $m$-th term of the sequence in $H$, and $x_n^{(m)}$ for its component in $H_n$.
Since $\lVert x_n^{(m)} - x_n^{(k)}\rVert_{H_n} \leqslant \lVert x^{(m)} - x^{(k)}\rVert_H$ for all $n$, we see that for a Cauchy sequence $(x^{(m)})_{m\in\mathbb{N}}$ in $H$, the sequences $(x_n^{(m)})_{m\in\mathbb{N}}$ of components are all Cauchy sequences in the respective $H_n$. By assumption, $H_n$ is complete, whence $\xi_n = \lim\limits_{m\to\infty} x_n^{(m)}$ exists for all $n$.
Next we see that $\xi = \{\xi_n\}_{n\in\mathbb{N}}$ is in $H$: Cauchy sequences are bounded, hence there is a $C < +\infty$ with $\lVert x^{(m)}\rVert \leqslant C$ for all $m$. Thus for all $N \in \mathbb{N}$ we have
$$\sum_{n=0}^N \lVert x_n^{(m)}\rVert_{H_n}^2 \leqslant C^2\tag{1}$$
for all $m \in\mathbb{N}$. Since $x_n^{(m)} \xrightarrow{m\to\infty} \xi_n$ implies $\lVert x_n^{(m)}\rVert_{H_n} \to \lVert \xi_n\rVert_{H_n}$, it follows that
$$\sum_{n=0}^N \lVert\xi_n\rVert_{H_n}^2 \leqslant C^2\tag{2}$$
by taking the limit of the finitely many summands in $(1)$. Since $N$ was arbitrary, it follows that
$$\sum_{n=0}^\infty \lVert \xi_n\rVert_{H_n}^2 = \lim_{N\to\infty} \sum_{n=0}^N \lVert\xi_n\rVert_{H_n}^2 \leqslant C^2,\tag{3}$$
and hence $\xi \in H$.
Finally, we see that $x^{(m)} \to \xi$ in $H$: Let $\varepsilon > 0$ be given. Since $(x^{(m)})$ is a Cauchy sequence, there is an $M \in \mathbb{N}$ such that
$$\lVert x^{(m)} - x^{(k)}\rVert \leqslant \varepsilon$$
for all $k,m \geqslant M$. For an arbitrary $m \geqslant M$ and all $N\in \mathbb{N}$ we then have
$$\sum_{n=0}^N \lVert x_n^{(m)} - \xi_n\rVert_{H_n}^2 = \lim_{k\to\infty} \underbrace{\sum_{n=0}^N \lVert x_n^{(m)} - x_n^{(k)}\rVert_{H_n}^2}_{\leqslant \lVert x^{(m)} - x^{(k)}\rVert^2} \leqslant \varepsilon^2,\tag{4}$$
and taking the limit for $N\to\infty$ in $(4)$ yields
$$\lVert x^{(m)} - \xi\rVert \leqslant \varepsilon$$
for all $m \geqslant M$, establishing
$$\xi = \lim_{m\to\infty} x^{(m)}$$
in $H$ and thereby the completeness of $H$.
A: I think you wanna use the following theorem:
A normed linear space $X$ is complete if and only if every absolutely summable series is summable.(Page 124 proposition 5 of chapter 6 in Real analysis by Royden, "Third Edition")
Due to that you first considered the absolutely summable series.
Set
$$H=:\oplus_{n=1}^\infty H_n=:\{\{h_n\}_n: \forall n\in\mathbb{N},\quad h_n\in H_n,\quad \|\{h_n\}_n\|_H=:\sum_{n=1}^\infty\|h_n\|_n^2<\infty \}$$
where $\|.\|_n$ is related norm to the $H_n$and suppose that $\{\{h_{n,m}\}_n\}_m$ is a absolutely summable series of elements of $H$ which means $\sum_{m=1}^\infty \|\{h_{n,m}\}_n\|_H<\infty$ . So we have
\begin{align}
\sum_{m=1}^\infty\sum_{n=1}^\infty(h_{n,m},h_{n,m})=\sum_{m=1}^\infty\sum_{n=1}^\infty \big((h_{n,m},h_{n,m})^\frac{1}{2}\big)^2=\sum_{m=1}^\infty\sum_{n=1}^\infty \|h_{n,m}\|_n^2=\sum_{m=1}^\infty \|\{h_{n,m}\}_n\|_H<\infty.
\end{align}
Now you only need to prove $\|\sum_{m=1}^\infty\{h_{m,n}\}_n\|_H<\infty$ or equivalently
$$\|\sum_{m=1}^\infty\{h_{m,n}\}_n\|_H=\|\{\sum_{m=1}^\infty h_{n,m}\}_n\|_H=\sum_{n=1}^\infty\|\sum_{m=1}^\infty h_{n,m}\|_n^2=\sum_{n=1}^\infty(\sum_{m=1}^\infty h_{n,m},\sum_{m=1}^\infty h_{n,m})<\infty$$
which will show $\{\{h_{n,m}\}_n\}_m$ is a summable series.(Notice that for all $n\in\mathbb{N},\quad \sum_{m=1}^\infty h_{m,n}\in H_n$)
It proves the completeness and that is the end.
A: Here the proof should be similar to how you show $\mathbb{R}$ is complete. Your proof got stuck because you did not use the condition
$$
|x|^{2}=\sum^{\infty}_{i=1}\langle x_{i},x_{i}\rangle<\infty
$$
